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\title{CHAPTER 12: TENSOR PRODUCTS}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
All the operations on $A_n$-modules to be defined in the next chapters make use of the tensor product, which we are about to study. 

The construction of the tensor product presented in \S 2 usually seems artificial on a first encounter. 

Fear not; it is the neat universal property of \S 3 that is most often used in the applications. 

The final two sections before the exercises contain a number of results that will be required later.

\end{frame}

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% Section 1
\section{BIMODULES.}
\begin{frame}[allowframebreaks]{A. }

To discuss tensor products in sufficient generality it is necessary to introduce bimodules. 

Let $R$ and $S$ be rings and let $M$ be an abelian group. 

To qualify as an $R$-$S$-bimodule, the group $M$ must be a left $R$-module and a right $S$-module, and the $R$-action and the $S$-action must be compatible in the sense that if $r \in R$, $s \in S$ and $u \in M$, then
\[ r(us) = (ru)s. \]

If $S = R$ then we simply say that $M$ is an $R$-bimodule.

A few examples and counter-examples will make the definition clear. 

If $R$ is a ring then $R^k$ is an $R$-bimodule. 

More generally, if $R$ is a subring of $S$, then $S^k$ is an $S$-$R$ bimodule. 

In particular this applies to Weyl algebras; for if $m < n$, then $A_m$ is a subring of $A_n$. 

Thus $A_n^k$ is an $A_n$-$A_m$-bimodule.

On the other hand, $K[X]$ is a left $A_1$-module and a right $K[X]$-module, but it is {\color{red}not} an $A_1$-$K[X]$-bimodule because the two actions are not compatible. 

Indeed, suppose that $f \in K[X]$ is acted on the left by $\partial$ and on the right by $x$. 

Then
\[ \partial(fx) = x\partial(f) + f \]
but $(\partial \cdot f)x = x\partial(f)$. 

More generally, $K[X]$ is {\color{red}not} an $A_n$-$K[X]$-bimodule.

Let $M$ be an $R$-$S$-bimodule. 

A subgroup $N$ of $M$ is a sub-bimodule of $M$ if it is stable under both the $R$-action and the $S$-action on $M$. 

In this case, the quotient group $M/N$ has a natural structure of $R$-$S$-bimodule.


The following example will often come up in applications. 

Consider $A_n$ as a subring of $A_{n+1}$ in the usual way. 

The left ideal $A_{n+1} x_{n+1}$ is a sub-bimodule of the $A_{n+1}$-$A_n$-bimodule $A_{n+1}$. 

Thus the quotient
\[ A_{n+1}/A_{n+1}x_{n+1} \]
is an $A_{n+1}$-$A_n$-bimodule.

We may similarly define a homomorphism of $R$-$S$-bimodules as a homomorphism of the underlying abelian groups which preserves both the left and right module structures.


\end{frame}

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% Section 2
\section{TENSOR PRODUCTS.}
\begin{frame}[allowframebreaks]{B. }

Let $R$, $S$ and $T$ be rings. 

Let $M$ be an $R$-$S$-bimodule and let $N$ be an $S$-$T$-bimodule. 

We will define the tensor product of $M$ and $N$ over $S$, denoted by $M \otimes_S N$.

First consider the set $M \times N$ of all pairs $(u,v)$, with $u \in M$, $v \in N$. 

Let $\mathcal{A}$ be the free abelian group whose basis is formed by the elements of $M \times N$. 

The elements of $\mathcal{A}$ are formal (finite) sums of the form
\[ \sum_i a_i(u_i,v_i) \]
with $a_i \in \mathbb{Z}$, $u_i \in M$, $v_i \in N$. 

Note that in this sum the pairs are mere symbols: the sum is not an element of the direct sum $M \oplus N$. 

In fact, if we assume that different indices correspond to different pairs in (2.1), then the sum is zero if and only if each $a_i = 0$. If $r \in R$ and $t \in T$, put
\[ r(u,v) = (ru,v), \]
\[ (u,v)t = (u,vt), \]
where $u \in M$ and $v \in N$. 

These are well-defined actions that make an $R$-$T$-bimodule of $\mathcal{A}$. 

Now consider the subgroup $\mathcal{B}$ of $\mathcal{A}$ generated by elements of the following types:
\[ (u+u',v) - (u,v) - (u',v), \]
\[ (u,v+v') - (u,v) - (u,v'), \]
\[ (us,v) - (u,sv), \]
where $r \in R$, $s \in S$, $t \in T$, $u,u' \in M$, $v,v' \in N$. 

One immediately checks that $\mathcal{B}$ is a sub-bimodule of $\mathcal{A}$.

The tensor product $M \otimes_S N$ is the quotient $\mathcal{A}/\mathcal{B}$. 

The image of $(u,v)$ in $M \otimes_S N$ is denoted by $u \otimes v$. 

Since $\mathcal{A}$ has the pairs $(u,v)$ as basis, the elements of the form $u \otimes v$ generate $M \otimes_S N$. 

However, it is {\color{red}not} true that every element of the tensor product is of the form $u \otimes v$.

The elements $u \otimes v$ satisfy relations which are determined by the generators of $\mathcal{B}$, as follows
\[ (u+u') \otimes v = u \otimes v + u' \otimes v, \]
\[ u \otimes (v+v') = u \otimes v + u \otimes v', \]
which are a sort of distributivity; and also
\[ us \otimes v = u \otimes sv, \]
which means that the tensor product is balanced; finally
\[ r(u \otimes v) = ru \otimes v, \]
\[ (u \otimes v)t = u \otimes vt, \]
are, respectively, the left $R$-action and right $T$-action on $M \otimes_S N$.

Note that if $R$ is a $K$-algebra then a left $R$-module may be seen as an $R$-$K$-bimodule. 

This trivial observation must be kept in mind in the applications. 

For example, if $M$ is a left $R$-module and $N$ a right $R$-module, then $N \otimes_R M$ is a $K$-vector space. 

We may also calculate $M \otimes_K N$, which is an $R$-bimodule. 

On the other hand, if $R$ is commutative then an $R$-module is automatically an $R$-bimodule. 

In particular if $M$ and $N$ are $R$-modules and $R$ is commutative, then $M \otimes_R N$ is an $R$-module.

Let us end this section by calculating a tensor product using the definition above. 

This is a simple example which, at the same time, carries a warning: the tensor product can behave in very unexpected ways. 

Consider the $\mathbb{Z}$-bimodules $\mathbb{Z}_p$ and $\mathbb{Z}_q$, and assume that $p$ and $q$ are co-prime. 

We wish to calculate $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$. 

We know that it is generated by elements of the form $m \otimes n$, where $m \in \mathbb{Z}_p, n \in \mathbb{Z}_q$. 

But since $p$ and $q$ are co-prime, there exist integers $a,b$ such that $ap + bq = 1$. 

Thus
\[ m \otimes n = (ap + bq)m \otimes n = bq \cdot m \otimes n \]
because $pm = 0$. 

Since the product is balanced, we deduce from the above that $m \otimes n = bm \otimes qn$. 

But $qn = 0$, and so we end up with $m \otimes n = 0$. 

Therefore, $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q = 0$.


\end{frame}

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% Section 3
\section{THE UNIVERSAL PROPERTY.}
\begin{frame}[allowframebreaks]{C. }

The first property of the tensor product that we shall study is also the key to proving all the other ones. 

To make its statement more succinct it is convenient to introduce some terminology. 

For the rest of this section let $R$, $S$ and $T$ be rings; let $M$ be an $R$-$S$-bimodule and $N$ an $S$-$T$-bimodule. 

Suppose that $L$ is an $R$-$T$-bimodule and that $\phi : M \times N \longrightarrow L$ is a map. 

We say that $\phi$ is bilinear if
\[ \phi(ru + u',v) = r\phi(u,v) + \phi(u',v), \]
\[ \phi(u,vt + v') = \phi(u,v)t + \phi(u,v'), \]
for all $u,u' \in M$, $v,v' \in N$, $r \in R$, $t \in T$. 

The map $\phi$ is said to be balanced, if
\[ \phi(us,v) = \phi(u,sv) \]
for all $s \in S$, $u \in M$, $v \in N$. 

The canonical example of a bilinear and balanced map is the projection
\[ \pi : M \times N \longrightarrow M \otimes_S N \]
defined by $\pi(u,v) = u \otimes v$.

Let $U$ be an $R$-$T$-bimodule, and suppose that there is a bilinear and balanced map $\eta : M \times N \longrightarrow U$. 

We say that $\eta$ is a universal bilinear balanced map if given any $R$-$T$-bimodule $L$ and a bilinear and balanced map $\phi : M \times N \longrightarrow L$, there exists a unique $R$-$T$-bimodule homomorphism $\overline{\phi} : U \longrightarrow L$ such that $\phi = \overline{\phi} \cdot \eta$.


\textbf{3.1. THEOREM.} The map $\pi : M \times N \longrightarrow M \otimes_S N$ is a universal bilinear balanced map.

\textbf{PROOF:} First of all, we may extend $\phi$ to get a map $\psi : \mathcal{A} \longrightarrow L$. In detail,
\[ \psi(\sum_i a_i (u_i, v_i)) = \sum_i a_i \phi(u_i, v_i). \]

Since $\phi$ is bilinear and balanced, it follows that $\psi$ is zero on every element of the submodule $\mathcal{B}$. Thus $\psi$ induces an $R$-$T$-bimodule map
\[ M \otimes_S N = \mathcal{A}/\mathcal{B} \longrightarrow L. \]

This is the map that we call $\overline{\phi}$. 

Since the canonical projection of $\mathcal{A}$ onto $\mathcal{A}/\mathcal{B}$ is induced by $\pi$, the equation $\phi = \overline{\phi} \cdot \pi$ follows immediately. 

We must now show that $\overline{\phi}$ is unique. 

Note that any bilinear and balanced map $\theta : M \otimes_S N \longrightarrow L$ for which $\phi = \theta \cdot \pi$ satisfies $\theta(u \otimes v) = \phi(u,v)$, for any $u \in M$, $v \in N$. 

But this equation is enough to characterize $\overline{\phi}$; hence $\theta = \overline{\phi}$.

It follows from this theorem that the universal property uniquely characterizes the tensor product.

\textbf{3.2. COROLLARY.} Let $\eta : M \times N \longrightarrow U$ be a universal bilinear balanced map. 

Then $U \cong M \otimes_S N$ as $R$-$T$-bimodules.

\textbf{PROOF:} Since $\pi$ is a universal bilinear balanced map, there exists $\overline{\eta} : M \otimes_S N \longrightarrow U$ such that $\eta = \overline{\eta} \cdot \pi$. 

On the other hand, since $\eta$ is also universal, there exists $\overline{\pi} : U \longrightarrow M \otimes_S N$ such that $\pi = \overline{\pi} \cdot \eta$. 

Thus $\eta = (\overline{\eta} \cdot \overline{\pi}) \cdot \eta$. 

Using again the universality of $\eta$, this time with respect to itself, and its uniqueness, we get that $\overline{\eta} \cdot \overline{\pi} = \iota$, where $\iota : U \longrightarrow U$ is the identity map. 

Similarly, $\overline{\pi} \cdot \overline{\eta}$ is the identity on $M \otimes_S N$.

We must now consider what happens with maps when we take tensor products. 

Let $M{\,}'$ be an $R$-$S$-bimodule and $N'$ be an $S$-$T$-bimodule. 

Let $\phi : M \longrightarrow M{\,}'$ and $\psi : N \longrightarrow N'$ be two bimodule homomorphisms. 

Consider the map $\theta : M \times N \longrightarrow M{\,}' \otimes_S N'$ defined by $\theta(u,v) = \phi(u) \otimes \psi(v)$. 

It follows from the properties of the tensor product that this map is bilinear and balanced. 

Hence, by the universal property of bilinear balanced maps, there exists a bimodule homomorphism:
\[ \overline{\theta} : M \otimes_S N \longrightarrow M{\,}' \otimes_S N'. \]

This map is usually denoted by $\phi \otimes \psi$. Note that
\[ (\phi \otimes \psi)(u \otimes v) = \phi(u) \otimes \psi(v). \]

\end{frame}

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% Section 4
\section{BASIC PROPERTIES.}
\begin{frame}[allowframebreaks]{D. }

We now use the universal property to prove some of the basic properties of the tensor product of modules. 

We illustrate the method with some examples, and leave the others as exercises for the reader.

\textbf{4.1. PROPOSITION.}
Let $R$, $S$ be rings, and $M$ be an $R$-$S$-bimodule. Then

\[ R \otimes_R M \cong M \cong M \otimes_S S. \]

\textbf{PROOF:} Let $\theta$ be the map from $R \times M$ to $M$ defined by $\theta(r,u) = ru$. 

It follows from the definition of an $R$-module that $\theta$ is bilinear and balanced. 

By the universal property of the tensor product there exists a bimodule homomorphism $\overline{\theta}$ from $R \otimes_R M$ to $M$. 

This map satisfies the equation $\overline{\theta}(r \otimes u) = ru$. 

It is clear that $\overline{\theta}$ is onto, we show that it is injective. 

Since we are tensoring up over $R$, we have that $r \otimes u = 1 \otimes ru$. 

Thus every element of $R \otimes_R M$ can be written in the form $1 \otimes u$, for some $u \in M$. 

But $\overline{\theta}(1 \otimes u) = u$, hence $\overline{\theta}$ must be injective. 

The proof of the other isomorphism is analogous and is left to the reader.

The property that we now prove is the distributivity of the tensor product with respect to the direct sum. It will be used often in later chapters.

\textbf{4.2. THEOREM.}
Let $R$, $S$, $T$ be rings, and $\mathcal{I}$ be an index set. 

Suppose that $M$ is an $R$-$S$-bimodule and that $N_i$ are $S$-$T$-bimodules, for every $i \in \mathcal{I}$. Then

\[ M \otimes_S \bigoplus_i N_i \cong \bigoplus_i (M \otimes_S N_i). \]

\textbf{PROOF:} The elements of $\bigoplus_i N_i$ will be denoted by $\bigoplus_i v_i$, where $v_i \in N_i$ for every $i \in \mathcal{I}$. 

Consider the map
\[ \theta : M \times \bigoplus_i N_i \longrightarrow \bigoplus_i (M \otimes_S N_i) \]
defined by $\theta(u,\bigoplus_i v_i) = \bigoplus_i (u \otimes v_i)$. 

It is a bilinear and balanced map. 

By the universal property of the tensor product, there exists a bimodule map
\[ \overline{\theta} : M \otimes_S \bigoplus_i N_i \longrightarrow \bigoplus_i (M \otimes_S N_i) \]
which satisfies $\overline{\theta}(u \otimes (\bigoplus_i v_i)) = \bigoplus_i (u \otimes v_i)$.

On the other hand, let $\mu_j : N_j \rightarrow \bigoplus_i N_i$ be the natural embedding. 

Thus we have maps
\[ Id \otimes \mu_j : M \otimes N_j \longrightarrow M \otimes_S \bigoplus_i N_i. \]

By the universal property of direct sums [Cohn 84, p.311], there exists a unique map
\[ \mu : \bigoplus_i (M \otimes N_i) \longrightarrow M \otimes (\bigoplus_i N_i) \]
such that $\mu(\bigoplus_i (u \otimes v_i)) = u \otimes (\bigoplus_i v_i)$. 

One now checks that $\mu$ and $\overline{\theta}$ are inverse to each other by a simple calculation on generators; hence both maps are isomorphisms.

A similar result holds when the direct sum is the first entry of the tensor product. 

Thus the following corollary is an immediate consequence of Theorem 4.2.

\textbf{4.3. COROLLARY.}
Let $R$ be a commutative ring and $\mathcal{I}_1$, $\mathcal{I}_2$ be index sets. 

Let $F_i$ be free $R$-modules with bases $\{v_\alpha^i : \alpha \in \mathcal{I}_i\}$, for $i=1,2$. 

Then $F_1 \otimes_R F_2$ is free with basis $\{v_\alpha^1 \otimes v_\beta^2 : \alpha \in \mathcal{I}_1, \beta \in \mathcal{I}_2\}$. 

Furthermore, if $F_1$ and $F_2$ have finite bases, then
\[ \operatorname{rank}(F_1 \otimes_R F_2) = \operatorname{rank}(F_1)\operatorname{rank}(F_2). \]

Let us now state a few more properties of the tensor product: the proofs follow the general pattern of 4.1 and 4.2.

\textbf{4.4. PROPOSITION.}
Let $R_i$ be rings and $M_i$ be $R_i$-$R_{i+1}$-bimodules, for $i=1,2,3$. 

Then
\[ M_1 \otimes_{R_2} (M_2 \otimes_{R_3} M_3) \cong (M_1 \otimes_{R_2} M_2) \otimes_{R_3} M_3. \]

\textbf{4.5. PROPOSITION.} Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. 

Then
\[ M \otimes_R N \cong N \otimes_R M. \]

We must also consider the effect of the tensor product on exact sequences. 

The first result holds in great generality.

\textbf{4.6. THEOREM.} Let $R$, $S$ and $T$ be rings and let 
\[ M{\,}' \xrightarrow{\psi} M \xrightarrow{\phi} M{\,}'' \to 0 \]
be an exact sequence of $S$-$T$-bimodules. 

If $B$ is an $R$-$S$-bimodule, then the sequence
\[ B \otimes_S M{\,}' \xrightarrow{1 \otimes \psi} B \otimes_S M \xrightarrow{1 \otimes \phi} B \otimes_S M{\,}'' \to 0 \]
is exact.

\textbf{PROOF:} Since $\phi$ is surjective, the elements of $B \otimes_S M{\,}''$ can be written in the form
\[ \sum_{1}^{k} b_i \otimes \phi(u_i) \]
where $b_i \in B$ and $u_i \in M$. Since this is equal to
\[ (1 \otimes \phi) \left( \sum_{1}^{k} b_i \otimes u_i \right) \]
we conclude that $1 \otimes \phi$ is surjective. 

On the other hand,
\[ (1 \otimes \phi) \cdot (1 \otimes \psi) = (1 \otimes \phi \cdot \psi) \]
equals zero, since $\ker(\phi) = \operatorname{im}(\psi)$. 

Hence
\[ \operatorname{im}(1 \otimes \psi) \subseteq \ker(1 \otimes \phi) \]
and the proof will be complete if we show that the opposite inclusion holds.

Let $N = \operatorname{im}(1 \otimes \psi)$. 

Then $\phi$ induces a map,
\[ \theta : (B \otimes_S M)/N \to B \otimes_S M{\,}'' \]
defined by $\theta(b \otimes u + N) = b \otimes \phi(u)$. 

Note that $\theta$ is well-defined because $N \subseteq \ker(1 \otimes \phi)$. Now let $\pi$ be the projection
\[ \pi : B \otimes_S M \to (B \otimes_S M)/N. \]

One easily checks that $\theta \cdot \pi = 1 \otimes \phi$. 

Suppose we have shown that $\theta$ is an isomorphism, then
\[ \ker(1 \otimes \phi) = \ker(\theta \cdot \pi) = \ker(\pi). \]

But $\ker(\pi) = N = \operatorname{im}(1 \otimes \psi)$, by definition. 

Thus the proof will be complete if we show that $\theta$ is an isomorphism. 

We do this by explicitly constructing an inverse.

Define a map,
\[ \beta : B \times M{\,}'' \to (B \otimes_S M)/N, \]
by $\beta(b,v) = b \otimes u + N$, where $\phi(u) = v$. 

We must check that it is well-defined. 

Assume that $\phi(u) = \phi(u') = v$, where $u,u' \in M$. 

Then $\phi(u-u') = 0$ and so
\[ u - u' \in \ker(\phi) = \operatorname{im}(\psi). \]

But we have that
\[ b \otimes u \equiv b \otimes u' \pmod{N} \]
and $\beta$ is well-defined. 

On the other hand, $\beta$ is clearly bilinear and $S$-balanced. 

By the universal property of the tensor product there exists a homomorphism of $R$-$T$-bimodules,
\[ \overline{\beta} : B \otimes_S M{\,}'' \to (B \otimes_S M)/N. \]

A straightforward calculation shows that $\theta$ and $\beta$ are inverse to each other.

Of course there is a version of Theorem 4.6 with $B$ tensoring on the right, and we shall use it whenever necessary without any further comment. 

The same applies to the next result.


\textbf{4.7. COROLLARY.} Let $S$ be a $K$-algebra and $R$ a subalgebra of $S$. 

Suppose that $I \subseteq S$ is an $S$-$R$-bimodule and that $M$ is a right $S$-module. 

Then
\[ M \otimes_S S/I \cong M/MI \]
as right $R$-modules.

\textbf{PROOF:} Tensoring the sequence $I \xrightarrow{\theta} S \to S/I \to 0$ with $M$ over $S$, and using Theorem 4.6, we obtain the exact sequence
\[ M \otimes_S I \xrightarrow{1 \otimes \theta} M \otimes_S S \to M \otimes S/I \to 0. \]

But $M \otimes_S S \cong M$ by Proposition 4.1. 

The image of the composition of $1 \otimes \theta$ with this isomorphism is $MI$. 

Note that this is a right $R$-submodule of $M$. 

Thus,
\[ M \otimes_S S/I \cong M/MI \]
as right $R$-modules.

An important warning: injective maps need {\color{red}not} be preserved by tensor products. 

Consequently, tensor products do {\color{red}not} always preserve short exact sequences. 

An example is given in Exercise 6.6. See also Exercise 6.5.

\end{frame}

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% Section 5
\section{LOCALIZATION.}
\begin{frame}[allowframebreaks]{E. }

It is time for our first application of the tensor product to the Weyl algebras. 

In it we generalize a result proved in Ch.10, §3. 

Let $p$ be a non-zero polynomial in $K[X]$ and consider the subset $K[X,p^{-1}]$ of all rational functions whose denominator is a power of $p$. 

We have proved that this is a {\color{red}holonomic} left $A_n$-module in Theorem 10.3.2.

Let $M$ be a $K[X]$-module and put
\[ M[p^{-1}] = K[X,p^{-1}] \otimes_{K[X]} M. \]

Since $K[X]$ is commutative, this is a $K[X]$-module. 

Note that every element of $M[p^{-1}]$ can be written in the form $p^{-k} \otimes u$, for some $u \in M$ and $k \geq 0$, but this form is not unique. 

This poses the question: when is $p^{-k} \otimes u = 0$?


\textbf{5.1. PROPOSITION.} Let $M$ be a $K[X]$-module. 

Suppose that $u \in M$ satisfies $1 \otimes u = 0$ in $M[p^{-1}]$. 

Then there exists an integer $k \geq 0$ such that $p^k u = 0$.

\textbf{PROOF:} There exist an index set $\mathcal{I}$ and a surjective map
\[ K[X]^{\mathcal{I}} \xrightarrow{\theta} M, \]
where $K[X]^{\mathcal{I}}$ denotes the direct sum of copies of $K[X]$ indexed by $\mathcal{I}$: a free $K[X]$-module. 

Let $F$ be the kernel of this map. 

Tensoring by $K[X,p^{-1}]$ over $K[X]$, we get an exact sequence,
\begin{equation}\tag{5.2}
K[X,p^{-1}] \otimes F \to K[X,p^{-1}] \otimes K[X]^{\mathcal{I}} \xrightarrow{1 \otimes \theta} K[X,p^{-1}] \otimes M \to 0.
\end{equation}

By Theorem 4.2,
\[ K[X,p^{-1}] \otimes_{K[X]} K[X]^{\mathcal{I}} \cong K[X,p^{-1}]^{\mathcal{I}}. \]

It is easy to check that the image of the first map of (5.2) in $K[X,p^{-1}]^{\mathcal{I}}$ is $K[X,p^{-1}]F$, the submodule generated by $F$. 

Therefore
\[ M[p^{-1}] \cong K[X,p^{-1}]^{\mathcal{I}} / K[X,p^{-1}]F \]
as $K[X]$-modules. Choose $v \in K[X]^{\mathcal{I}}$ such that $\theta(v) = u$. 

Thus
\[ (1 \otimes \theta)(1 \otimes v) = 1 \otimes u = 0. \]

Identifying $1 \otimes v$ with $v$ in $K[X,p^{-1}]^{\mathcal{I}}$, we have that $v \in K[X,p^{-1}]F$. 

Hence there exists $k \geq 0$ such that $p^k v \in F$; from which we conclude that $p^k u = 0$ in $M$.

Now assume that $M$ is a left $A_n$-module. 

Then $M[p^{-1}]$ is a $K[X]$-module, and we want to turn it into a left $A_n$-module. 

We do that by defining the action of $\partial_i$ as follows
\[ \partial_i \cdot (p^{-k} \otimes u) = p^{-k-1} \otimes (p \partial_i \cdot u - k \partial_i(p) u). \]

It is not immediately clear that this formula gives a well-defined action, because an element of $M[p^{-1}]$ does not have a unique representation in the
form $p^{-k} \otimes u$. 

However, applying Proposition 5.1 one can show that it is indeed well-defined; the details are left to the reader. 

It is also easy to check that the relations of $A_n$ are preserved by this action. 

Hence $M[p^{-1}]$ is a left $A_n$-module by Appendix 1. 

It is called the {\color{red}localization} of $M$ at $p$. 

The name comes from the way this construction is used in algebraic geometry.

If we also assume that $M$ has a good filtration $\Gamma$, then we can construct a filtration $\Omega$ for $M[p^{-1}]$. 

If $m$ is the degree of $p$, put
\begin{equation}\tag{5.3}
\Omega_k = \{ p^{-k} \otimes u : u \in \Gamma_{(m+1)k} \}.
\end{equation}

To check that this is a filtration of $M[p^{-1}]$, proceed as in the proof of Theorem 10.3.2.

\textbf{5.4. THEOREM.} If $M$ is a {\color{red}holonomic} $A_n$-module, then so is $M[p^{-1}]$.

\textbf{PROOF:} We have that
\[ \dim_K \Omega_k \leq \dim_K \Gamma_{(m+1)k}. \]
Let $\chi(t,M,\Gamma)$ be the Hilbert polynomial of $\Gamma$. 

Then for $k \gg 0$,
\[ \dim_K \Omega_k \leq \chi(k(m+1), M, \Gamma) \]
a polynomial of degree $n$. Hence $M[p^{-1}]$ is {\color{red}holonomic} by Lemma 10.3.1.

\end{frame}

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% Section 6
\section{EXERCISES.}
\begin{frame}[allowframebreaks]{F. }

\begin{enumerate}

\item \textbf{6.1. Exercise.} Let $p,q \in \mathbb{Z}$, and let $m$ be their greatest common divisor. 

Show that $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q \cong \mathbb{Z}_m$.

\newpage

\item \textbf{6.2. Exercise.} Prove the following isomorphisms of $\mathbb{Z}$-modules:
\begin{enumerate}
    \item[(1)] $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$.
    \item[(2)] $\mathbb{Q} \otimes_{\mathbb{Z}} (\mathbb{Q}/\mathbb{Z}) \cong (\mathbb{Q}/\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Q}/\mathbb{Z})$.
\end{enumerate}

\newpage

\item \textbf{6.3. Exercise.} Let $R$ be a ring. 

Show that $R^n \otimes_R R^m \cong R^{nm}$.

\newpage

\item \textbf{6.4. Exercise.} Prove Propositions 4.4 and 4.5 in detail.

\newpage

\item \textbf{6.5. Exercise.} Let $R$ be a $K$-algebra and let $k \geq 0$ be an integer. 

Show that if $N \to M$ is an injective map of left $R$-modules, then so is
\[ R^k \otimes_R N \to R^k \otimes_R M. \]

\newpage

\item \textbf{6.6. Exercise.} Let $\phi : \mathbb{Z}_2 \to \mathbb{Z}_4$ be the injective homomorphism of $\mathbb{Z}$-modules defined by $\phi(1 + 2\mathbb{Z}) = 2 + 4\mathbb{Z}$. 

Show that:
\begin{enumerate}
    \item[(1)] $\mathbb{Z}_2 \otimes_{\mathbb{Z}} \mathbb{Z}_2 \cong \mathbb{Z}_2$.
    \item[(2)] $\mathbb{Z}_2 \otimes_{\mathbb{Z}} \mathbb{Z}_4 \cong \mathbb{Z}_2$.
    \item[(3)] The map $\mathbb{Z}_2 \to \mathbb{Z}_2$ induced by tensoring $\phi$ by $\mathbb{Z}_2$ over $\mathbb{Z}$ is identically zero.
\end{enumerate}
Conclude that the tensor product by $\mathbb{Z}_2$ does {\color{red}not} preserve injectivity.

\newpage

\item \textbf{6.7. Exercise.} Show that for any non-zero $p \in K[X]$, there exist $A_n$-module isomorphisms:
\begin{enumerate}
    \item[(1)] $K[X,p^{-1}] \otimes_{K[X]} K[X] \cong K[X,p^{-1}]$;
    \item[(2)] $K[X,p^{-1}] \otimes_{K[X]} K[\partial_1, \ldots, \partial_n] = 0$.
\end{enumerate}

\end{enumerate}

\end{frame}

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\end{document}


